Vectors A⃗ and B⃗ lie in an x y plane. A⃗ has magnitude 8.00 and angle 130^∘ ; B⃗ has components

Vectors A⃗ and B⃗ lie in an x y plane. A⃗ has magnitude 8.00...

Vectors $\vec{A}$ and $\vec{B}$ lie in an $x y$ plane. $\vec{A}$ has magnitude 8.00 and angle $130^{\circ} ; \vec{B}$ has components $B_{x}=-7.72$ and $B_{y}=-9.20$ What are the angles between the negative direction of the $y$ axis and $(a)$ the direction of $\vec{A},$ (b) the direction of the product $\vec{A} \times \vec{B},$ and $(c)$ the direction of $\vec{A} \times(\vec{B}+3.00 \hat{k}) ?$

Vectors $\vec{A}$ and $\vec{B}$ lie in an $x y$ plane. $\vec{A}$ has magnitude 8.00 and angle $130^{\circ} ; \vec{B}$ has components $B_{x}=-7.72$ and $B_{y}=-9.20$ What are the angles between the negative direction of the $y$ axis and $(a)$ the direction of $\vec{A},$ (b) the direction of the product
$\vec{A} \times \vec{B},$ and $(c)$ the direction of $\vec{A} \times(\vec{B}+3.00 \hat{k}) ?$

Key Concepts

Vector Representation in the Coordinate Plane

This concept deals with expressing vectors using their components along the coordinate axes (i.e., x and y components in a plane). It involves understanding how to represent and analyze vectors graphically and numerically by breaking them down into components, which simplifies operations such as addition, subtraction, and the computation of angles between vectors.

Conversion Between Polar and Cartesian Coordinates

Often vectors are provided in polar form (magnitude and angle) and must be converted into Cartesian coordinates (x and y components) for further analysis. This conversion uses trigonometric relations (sine and cosine functions) to find the vector's components, which is critical when comparing vectors or computing cross products in a coordinate system.

Angle Determination Using Vector Components

Determining the angle a vector makes with respect to a particular axis involves using the components of the vector and inverse trigonometric functions. This process often requires adjustments depending on the specific reference direction (for example, the negative y-axis) and careful consideration of the vector’s quadrant to arrive at the correct directional angle.

Cross Product of Vectors

The cross product is an operation on two vectors in three-dimensional space that results in a new vector perpendicular to the plane containing the original vectors. Its direction follows the right-hand rule, and its magnitude depends on the sine of the angle between the original vectors. The concept is crucial when modifying or analyzing vector fields, especially for understanding directions perpendicular to a given plane.

Extension to Three Dimensions with Additional Components

Even when initial vectors lie in a plane, adding a component along the z-axis introduces three-dimensional considerations. This concept emphasizes understanding how the inclusion of an extra dimension affects operations such as the cross product, as well as how vector directions and angles are interpreted in a three-dimensional space.

Angle Measurement Relative to Non-standard Reference Axes

Often angles are measured relative to non-standard reference directions, such as the negative y-axis rather than the positive x-axis. This concept involves adjusting standard angle calculations to account for a different reference axis, requiring a clear understanding of coordinate geometry and angle conversion methods.

Transcript

00:02 So noting that the given angle is 130 degrees here and that the magnitude of the ax and ay for vector a is both 8, we can then find the i -hat and j -hat components for vector a.

00:16 Okay.

00:18 So vector a, the i -hat component is going to be equal to 8 times the cosine of 130 degrees, which is minus 5 .14 in the i -hat direction, plus 8 times.

00:36 The sign of 130 degrees, which is 6 .13.

00:41 And that's in the j -hat direction.

00:44 Then b, here, of course, is just going to be equal to b of x, which is minus 7 .72, in the i -hat direction, plus b of y, which is minus 9 .20 in the j -hat direction.

01:02 Okay.

01:03 Then for part a, to find the angle, the angle cosine of theta is equal to the adjacent, which here would be for this question part a is equal to a times a dot product of minus j hat divided by the hypotenuse which is just the magnitude of a okay well then solving for theta the angle it's going to be equal to the inverse cosine of that ratio well magnitude of a dot negative j would be um minus 6 .13.

01:50 So it's just the dot product is just the light components.

01:54 So that's the light component from a.

01:56 Then you have the minus sign still, of course.

01:58 Then the magnitude of a is going to be equal to the square root of the sum of the squares.

02:03 So this would be 5 .14 squared plus 6 .13 squared...

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